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# M02-21

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:18
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65% (hard)

Question Stats:

57% (01:48) correct 43% (02:05) wrong based on 342 sessions

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If $$x$$ is a positive integer, is $$y^2 (x^3 - x + 1) \gt 75?$$

(1) $$y \gt 1$$

(2) $$(x-1)^2=16$$

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16 Sep 2014, 00:18
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Official Solution:

If $$x$$ is a positive integer, is $$y^2 (x^3 - x + 1) \gt 75?$$

We need to know values of two variables to answer the question:
1. The value of $$x$$
2. The value of $$y$$

Statement (1) by itself is insufficient. We only know that $$y \gt 1$$.

Statement (2) by itself is insufficient. We know the value of $$x$$. Knowing that $$x$$ is a positive integer, we conclude that $$x-1 = 4$$, thus, $$x = 5$$. However, we don't know the value of $$y$$.

Statements (1) and (2) combined are sufficient. If $$x = 5$$, then $$5^3 - 5 + 1 = 121$$. If $$y \gt 1$$, then the statement will always be greater than 75.

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11 Oct 2014, 03:55
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from (2): we can also get the x=-3. Bunuel can i infer this?
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11 Oct 2014, 04:00
jacobneroth wrote:
from (2): we can also get the x=-3. Bunuel can i infer this?

No, because the stem says that x is a positive integer.
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11 Apr 2015, 08:06
Bunuel wrote:
Official Solution:

We need to know values of two variables to answer the question:
1. The value of $$x$$
2. The value of $$y$$

Statement (1) by itself is insufficient. We only know that $$y \gt 1$$.

Statement (2) by itself is insufficient. We know the value of $$x$$. Knowing that $$x$$ is a positive integer, we conclude that $$x-1 = 4$$, thus, $$x = 5$$. However, we don't know the value of $$y$$.

Statements (1) and (2) combined are sufficient. If $$x = 5$$, then $$x = 5^3 - 5 + 1 = 121$$. If $$y \gt 1$$, then the statement will always be greater than 75.

I'm not sure I fully understand this. If $$x = 5$$ then, then $$x = 5^3 - 5 + 1 = 121$$, as such, $$y^2 > 75/121$$ which is always true, whether $$y = 1$$ or $$y = -1$$. That should be the second statement of x=5 is sufficient.
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11 Apr 2015, 08:15
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gmatser1 wrote:
Bunuel wrote:
Official Solution:

We need to know values of two variables to answer the question:
1. The value of $$x$$
2. The value of $$y$$

Statement (1) by itself is insufficient. We only know that $$y \gt 1$$.

Statement (2) by itself is insufficient. We know the value of $$x$$. Knowing that $$x$$ is a positive integer, we conclude that $$x-1 = 4$$, thus, $$x = 5$$. However, we don't know the value of $$y$$.

Statements (1) and (2) combined are sufficient. If $$x = 5$$, then $$x = 5^3 - 5 + 1 = 121$$. If $$y \gt 1$$, then the statement will always be greater than 75.

I'm not sure I fully understand this. If $$x = 5$$ then, then $$x = 5^3 - 5 + 1 = 121$$, as such, $$y^2 > 75/121$$ which is always true, whether $$y = 1$$ or $$y = -1$$. That should be the second statement of x=5 is sufficient.

What if y=0? Or y=1/2? Or y=-1/2? There are infinitely many values of y, for which y^2>75/121 is not true.
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14 Apr 2015, 23:50
Yes even I had the same question , but if we read the question carefully it says x is a positive integer. So, x=-3 cannot be considered in this case.
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27 Feb 2016, 08:21
I think this is a high-quality question and I don't agree with the explanation. Hi,

Since y is squared in the statement, a negative integer will not matter, right? Hence, shouldn't statement 2 alone be sufficient?
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29 Feb 2016, 09:10
anisharao2 wrote:
I think this is a high-quality question and I don't agree with the explanation. Hi,

Since y is squared in the statement, a negative integer will not matter, right? Hence, shouldn't statement 2 alone be sufficient?

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22 Aug 2016, 14:09
I think this is a high-quality question and I agree with explanation.
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01 Dec 2016, 09:15
I think this is a high-quality question and I don't agree with the explanation. As per Statement 2 - X may have 2 possible solutions 5 and -3 so even if we analyse both statement together than also the whole value might not be greater than 75
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01 Dec 2016, 09:43
anujjain2016 wrote:
I think this is a high-quality question and I don't agree with the explanation. As per Statement 2 - X may have 2 possible solutions 5 and -3 so even if we analyse both statement together than also the whole value might not be greater than 75

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22 Mar 2017, 04:40
Hi Bunuel,

For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (-infinity to -(75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?
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22 Mar 2017, 05:14
akshayv7 wrote:
Hi Bunuel,

For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (-infinity to -(75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?

Not sure I understand what you mean...

The question asks whether $$y^2 (x^3 - x + 1) \gt 75$$, not whether y > 0. (2) says that y > 0. From (1) the question becomes whether $$y^2 \gt 75/121$$ and since from (2) we know that y > 0, then when considering the statements together we can answer that YES $$y^2 \gt 75/121$$.
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20 Jun 2017, 00:14
I solved statement 2 in following manner:

(x-1)^2=16
x^2-2x+1=16
x^2-2x=15
x(x-2)=15
So x=0 or x=17

Where did I go wrong?
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20 Jun 2017, 01:19
disharupani wrote:
I solved statement 2 in following manner:

(x-1)^2=16
x^2-2x+1=16
x^2-2x=15
x(x-2)=15
So x=0 or x=17

Where did I go wrong?

You can equate multiples to 0, when the whole product equals 0. For example, if it were x(x - 2) = 0, then yes, x = 0 or x = 2. Notice that if x = 0, then x(x - 2) = 15, does not hold. Also, it's not clear how you got x = 17, there. Again, if x = 17, then x(x - 2) = 15, does not hold.
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29 Aug 2017, 12:39
Hi Bunuel,

1 clarification needed..
in statement 2->
we get x=5
y^2 * 121>75 to be proven?
y^2 > or = to zero
even if we take y as 1/2 then y^2 =1/4
1/4 *121=30.25 <75 ---we get answer as NO

Put y=0 then y^2=0
0*121=0 <75 ---we get answer as NO

Put y=-12/11 then y^2=144/121
144/121 *121=144 >75 ---we get answer as YES

if y= integer then definitely y^2 * 121>75

Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.

Thanks,
Stuti
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29 Aug 2017, 21:45
1
stutimittal04 wrote:
Hi Bunuel,

1 clarification needed..
in statement 2->
we get x=5
y^2 * 121>75 to be proven?
y^2 > or = to zero
even if we take y as 1/2 then y^2 =1/4
1/4 *121=30.25 <75 ---we get answer as NO

Put y=0 then y^2=0
0*121=0 <75 ---we get answer as NO

Put y=-12/11 then y^2=144/121
144/121 *121=144 >75 ---we get answer as YES

if y= integer then definitely y^2 * 121>75

Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.

Thanks,
Stuti

For some values of y, the answer is YES, for example if y = 100.
For some values of y, the answer is NO, for example if y = 0.
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31 Aug 2017, 09:58
Ohok!! Thanks

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
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25 Oct 2017, 00:08
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What do you guys think about the variable approach from Math Revolution? I think that in some cases it is better to use this approach, but in most cases - not. And if you are aiming for, at least, Q50, variable approach should not be your friend
Re: M02-21   [#permalink] 25 Oct 2017, 00:08

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# M02-21

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